| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2022 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circles |
| Type | Two circles intersection or tangency |
| Difficulty | Standard +0.3 This is a standard circles intersection problem requiring the cosine rule to find angles, then applying arc length and sector area formulas. While it involves multiple steps (finding angle at centre, calculating two arc lengths, finding sector areas, subtracting triangle area), these are all routine techniques for P1 level with no novel insight required. The identical circles and given ratio simplify the problem considerably. |
| Spec | 1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.03f Circle properties: angles, chords, tangents1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta |
| Answer | Marks | Guidance |
|---|---|---|
| \(APQ = \cos^{-1}\frac{\frac{5}{6}r}{r} \left[= \cos^{-1}\frac{5}{6}\right]\) | \*M1 | May use cosine rule to find APB. Stating APQ or APB as an incorrect multiple of \(\pi\) is M0. |
| \(= 0.5857\) | A1 | Accept 0.586 or 33.6° or APB (1.171 or 67.1°). |
| Perimeter \(= 4 \times r \times their\ 0.5857 = 2.34r\) or \(0.745\pi r\) or \((293/125)r\) | DM1 A1 | Must use a numerical value of *their* angle. |
| Answer | Marks | Guidance |
|---|---|---|
| Use of sector formula: Sector APB \(= \frac{1}{2}r^2 \times (2 \times their\ 0.5857)\) or Sector APC (C is on PQ so PC \(= r\)) \(= \frac{1}{2}r^2 \times (their\ 0.5857)\) | M1 | Any sector with *their* appropriate angle. It must be clear the appropriate numerical angle is being used. |
| Use of appropriate formula for area of triangle and correct combination with the sector to find the area of a half segment, one segment or both segments | M1 | e.g. Area APB \(= \frac{1}{2}r^2 \times \sin(2 \times their\ 0.5857)\). |
| Shaded area \([= 2 \times 0.1250r^2] = 0.250r^2\) | A1 | or \(0.0796\pi r^2\), allow \(\frac{1}{4}r^2\) or \(0.25r^2\). |
## Question 8:
### Part 8(a):
$APQ = \cos^{-1}\frac{\frac{5}{6}r}{r} \left[= \cos^{-1}\frac{5}{6}\right]$ | **\*M1** | May use cosine rule to find APB. Stating APQ or APB as an incorrect multiple of $\pi$ is M0.
$= 0.5857$ | **A1** | Accept 0.586 or 33.6° or APB (1.171 or 67.1°).
Perimeter $= 4 \times r \times their\ 0.5857 = 2.34r$ or $0.745\pi r$ or $(293/125)r$ | **DM1 A1** | Must use a numerical value of *their* angle.
### Part 8(b):
Use of sector formula: Sector APB $= \frac{1}{2}r^2 \times (2 \times their\ 0.5857)$ or Sector APC (C is on PQ so PC $= r$) $= \frac{1}{2}r^2 \times (their\ 0.5857)$ | **M1** | Any sector with *their* appropriate angle. It must be clear the appropriate numerical angle is being used.
Use of appropriate formula for area of triangle and correct combination with the sector to find the area of a half segment, one segment or both segments | **M1** | e.g. Area APB $= \frac{1}{2}r^2 \times \sin(2 \times their\ 0.5857)$.
Shaded area $[= 2 \times 0.1250r^2] = 0.250r^2$ | **A1** | or $0.0796\pi r^2$, allow $\frac{1}{4}r^2$ or $0.25r^2$.
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\includegraphics[max width=\textwidth, alt={}, center]{8eb3d21b-dc45-493c-9e5c-3c0535c505e8-10_492_888_255_625}
The diagram shows two identical circles intersecting at points $A$ and $B$ and with centres at $P$ and $Q$. The radius of each circle is $r$ and the distance $P Q$ is $\frac { 5 } { 3 } r$.
\begin{enumerate}[label=(\alph*)]
\item Find the perimeter of the shaded region in terms of $r$.
\item Find the area of the shaded region in terms of $r$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2022 Q8 [7]}}